
\prob{0002}{40度角}

\begin{figure}[htbp]
  \centering
  \image{0002}
  \caption{0002：40度角} \label{fig:0002}
\end{figure}

如图~\ref{fig:0002}，$\angle C = 40^\circ, \angle BDC = 60^\circ, AB = CD$，求$\angle A$。
\problabels{yellow/平面几何, green/角度问题}

\ans{$\angle A = 30^\circ$}

\subsection{构造等边三角形} \label{subsec:0002-eqtri}

\begin{figure}[htbp]
  \centering
  \image{0002-eqtri}
  \caption{\nameref{subsec:0002-eqtri}：通过构造等边三角形找出一系列等腰三角形。} \label{fig:0002-eqtri}
\end{figure}

基本思路：构造等边三角形，然后通过一系列等腰三角形求解。

如图~\ref{fig:0002-eqtri}，延长$DB$到$E$使得$DE = CE$，$E$在直线$AC$下方。在直线$AC$上找到一点$F$使得$EF = EB$，连接$EA, EC, EF$。

\begin{align*}
  \because  {}& DC = DE, \angle CDE = 60^\circ \\
  \therefore{}& \text{$\triangle CDE$是等边三角形} \\
  \therefore{}& DC = DE = CE, \\
  & \angle CDE = \angle DCE = \angle DEC = 60^\circ \\
  \because  {}& AB = CD \\
  \therefore{}& AB = CE \\
  \because  {}& \angle ACD = 40^\circ \\
  \therefore{}& \angle ACE = 20^\circ \\
  \therefore{}& \angle ABE = 80^\circ \\
  \because  {}& EF = EB \\
  \therefore{}& \angle ABE = \angle CFE \\
  \therefore{}& \angle CFE = 80^\circ \\
  \therefore{}& \angle DEF = 20^\circ \\
  \therefore{}& \angle CEF = 80^\circ \\
  \therefore{}& \angle CFE = \angle CEF \\
  \therefore{}& \angle CEF = \angle ABE, CE = CF \\
  \because  {}& \text{在 $\triangle ABE$ 与 $\triangle CEF$ 中} \\
  & \begin{cases}
    AB = CE \\
    \angle ABE = \angle CEF \\
    BE = EF
  \end{cases} \\
  \therefore{}& \triangle ABE \cong \triangle CEF \\
  \therefore{}& AE = CF, \angle AED = \angle CFE, \angle EAC = \angle ECA \\
  \therefore{}& DC = DE = CE = CF = AB = AE \\
  \therefore{}& EA = ED \\
  \therefore{}& \angle EAD = \angle EDA \\
  \because  {}& \angle AED = 80^\circ \\
  \therefore{}& \angle EAD = 50^\circ \\
  \because  {}& \angle EAC = 20^\circ \\
  \therefore{}& \angle CAD = 30^\circ
\end{align*}

综上，$\angle CAD = 30^\circ$。
